Prove that the equation $\cos x = bx$ has a single solution in the closed interval $[0, \frac{\pi}{2}]$ if $b \ge 0$.
1) If $b = 0$
In such case $\cos x=0 \cdot x$ and only one $x$ can satisfy such equation in the given interval which is $\frac{\pi}{2}$.
2) If $b > 0$
Let $E(x) = \cos x - bx$
$E(0) > 0$
$E(\frac{\pi}{2}) < 0$
$E(x)$ is continuous for all $x$ as it is the difference of a polynomial and $\cos$. Thus according to intermediate value theorem there must be an $x_1$ in the interval $(0, \frac{\pi}{2})$ such that $E(x_1) = 0$. Suppose that there's another $x= x_2 \neq x_1$ such that $E(x_2) = 0$. $E(x)$ is continuous in $[0, \frac{\pi}{2}]$ and differentiable in $(0, \frac{\pi}{2})$ thus according to Rolle theorem there must a root to $E'(x)$.
$E'(x) = -\sin x - b$
If we try to find the root of $E'(x)$:
$-\sin x-b = 0$
$-\sin x = b$
But $\sin x>0$ in the interval $[0, \frac{\pi}{2}]$ and $b>0$ hence it's a contradiction. Thus there's not a second root for $E(x)$ in the open interval.
Now we just need to check if $x = 0$ or $x = \frac{\pi}{2}$ will result in a zero.
$\cos\frac{\pi}{2}=b\frac{\pi}{2}$
$0 = b\frac{\pi}{2} \Rightarrow$ contradiction as $b > 0$
$\cos 0 = b\cdot0$
$1 \neq 0 \Rightarrow$ contradiction.
Not sure if I actually had to do the last part of check the values on the edges of the closed interval, I think I have to because Intermidiate value and Rolle theorems talk about having results in the open interval.
Please let me know if my solution is correct.
